Question

Gỉai: \(\sqrt{2x^2+21x-11}-3\sqrt{2x-1}=\sqrt{2x^2-9x+4}\)

Answer 1

\(\sqrt{2x^2+21x-11}-3\sqrt{2x-1}=\sqrt{2x^2-9x+4}\)

\(\Leftrightarrow\sqrt{\left(2x-1\right)\left(x+11\right)}-3\sqrt{2x-1}-\sqrt{\left(2x-1\right)\left(x-4\right)}=0\)

\(\Leftrightarrow\sqrt{2x-1}\left(\sqrt{x+11}-3-\sqrt{x-4}\right)=0\)

☘ Trường hợp 1:

\(\sqrt{2x-1}=0\)

⇔ 2x - 1 = 0

⇔ x = 0,5 (nhận)

☘ Trường hợp 1:

\(\sqrt{x+11}-3-\sqrt{x-4}=0\)

\(\Leftrightarrow\sqrt{x-4}+3=\sqrt{x+11}\)

\(\Leftrightarrow x-4+6\sqrt{x-4}+9=x+11\)

\(\Leftrightarrow6\sqrt{x-4}=6\)

\(\Leftrightarrow\sqrt{x-4}=1\)

⇔ x - 4 = 4

⇔ x = 5 (nhận)

☘ Vậy \(S=\left\{\dfrac{1}{2};5\right\}\)